Question

8. For the Sine Waveform with a Value of 440VRms,what is the Peak Voltage (Vpk)?

9. What are the peak-to-peak voltage and frequency (in Hz ) of the wave v(t)=169 V sin(377t) ? 10. Find out the phase difference between the following waveforms:x₁(t)=10sin(ωt+120˚) and x₂(t)=20cos(ωt+150˚)

Answer 1

8. The peak voltage
`(\(V_{\text{pk}}\))` of a sine waveform with a value of 440
`V\(_{\text{RMS}}\) is \(V_{\text{pk}} = √(2) * V_{\text{RMS}} = √(2) * 440 \, \text{V} \approx 622.3 \, \text{V}\)`.

9. For the waveform
`\(v(t) = 169 \, \text{V} \sin(377t)\)`, the peak-to-peak voltage is
`\(V_{\text{pp}} = 2 * \text{Amplitude} = 2 * 169 \, \text{V} = 338 \, \text{V}\)`, and the frequency is
`\(f = (\omega)/(2\pi) = (377)/(2\pi) \, \text{Hz}\)`.

10. The phase difference between the waveforms
`\(x_1(t) = 10 \sin(\omega t + 120^\circ)\) and \(x_2(t) = 20 \cos(\omega t + 150^\circ)\) is \(30^\circ\) (or \((\pi)/(6)\) radians)`.

Step-by-step explanation:

8. The RMS (Root Mean Square) value of a sine waveform is related to its peak value by the formula
`\(V_{\text{RMS}} = \frac{V_{\text{pk}}}{√(2)}\)`. By rearranging this formula, we can find the peak voltage
`(\(V_{\text{pk}}\))` in terms of the RMS voltage. For a waveform with an RMS value of 440 V, the peak voltage is
`\(V_{\text{pk}} = √(2) * V_{\text{RMS}}\)`, yielding approximately 622.3 V.

9. The given waveform
`\(v(t) = 169 \, \text{V} \sin(377t)\)` can be analyzed to determine its peak-to-peak voltage and frequency. The peak-to-peak voltage is twice the amplitude of the waveform, which in this case is 169 V, resulting in a peak-to-peak voltage of 338 V. The frequency (f) of the waveform is calculated by using the angular frequency
`\(\omega = 377\)` rad/s and applying the formula
`\(f = (\omega)/(2\pi)\)`, giving
`\(f = (377)/(2\pi)\)` Hz.

10. To find the phase difference between two waveforms,
`\(x_1(t) = 10 \sin(\omega t + 120^\circ)\)` and
`\(x_2(t) = 20 \cos(\omega t + 150^\circ)\)`, we compare their phase angles. Converting the cosine function in
`\(x_2(t)\)` to a sine function with a phase shift, we have
`\(x_2(t) = 20 \sin(\omega t + 60^\circ)\)`. Now, comparing
`\(x_1(t)\)` and the modified
`\(x_2(t)\)`, we notice a
`\(60^\circ\)` phase difference between them. Therefore, the phase difference between the original
`\(x_1(t)\)` and
`\(x_2(t)\)` is
`\(60^\circ - 120^\circ = 30^\circ\)` (or
`\((\pi)/(6)\)` radians).